The foot of the perpendicular from (2,4,-1) to the line `(x-6)/(1)=(2-y)/(2)=(z-2)/(2)` is

Answer

To find the foot of the perpendicular from the point \( P(2, 4, -1) \) to the line given by the symmetric equations \( \frac{x-6}{1} = \frac{2-y}{2} = \frac{z-2}{2} \), we will follow these steps:

### Step 1: Write the parametric equations of the line
From the symmetric equations of the line, we can express the coordinates of any point \( Q \) on the line in terms of a parameter \( k \):
- \( x = 1k + 6 \) (or \( x = k + 6 \))
- \( y = 2 - 2k \)
- \( z = 2k + 2 \)

So, the coordinates of point \( Q \) can be written as:
\[ Q(k) = (k + 6, 2 - 2k, 2k + 2) \]

### Step 2: Find the direction ratios
The direction ratios of the line are given by the coefficients in the symmetric equations, which are \( (1, -2, 2) \).

### Step 3: Find the direction vector \( \overrightarrow{PQ} \)
The vector \( \overrightarrow{PQ} \) from point \( P(2, 4, -1) \) to point \( Q(k) \) is:
\[ \overrightarrow{PQ} = Q(k) - P = (k + 6 - 2, (2 - 2k) - 4, (2k + 2) - (-1)) \]
This simplifies to:
\[ \overrightarrow{PQ} = (k + 4, -2 - 2k, 2k + 3) \]

### Step 4: Set up the dot product equation
For the line and the vector \( \overrightarrow{PQ} \) to be perpendicular, their dot product must equal zero:
\[ (k + 4, -2 - 2k, 2k + 3) \cdot (1, -2, 2) = 0 \]
Calculating the dot product:
\[ (k + 4) \cdot 1 + (-2 - 2k)(-2) + (2k + 3) \cdot 2 = 0 \]
Expanding this gives:
\[ k + 4 + 4 + 4k + 4k + 6 = 0 \]
Combining like terms:
\[ 9k + 14 = 0 \]

### Step 5: Solve for \( k \)
Solving for \( k \):
\[ 9k = -14 \implies k = -\frac{14}{9} \]

### Step 6: Substitute \( k \) back into the equations for \( Q \)
Now, substitute \( k = -\frac{14}{9} \) back into the equations for \( Q \):
- \( x = -\frac{14}{9} + 6 = -\frac{14}{9} + \frac{54}{9} = \frac{40}{9} \)
- \( y = 2 - 2(-\frac{14}{9}) = 2 + \frac{28}{9} = \frac{18}{9} + \frac{28}{9} = \frac{46}{9} \)
- \( z = 2(-\frac{14}{9}) + 2 = -\frac{28}{9} + \frac{18}{9} = -\frac{10}{9} \)

### Final Answer
Thus, the foot of the perpendicular from the point \( (2, 4, -1) \) to the line is:
\[ \left( \frac{40}{9}, \frac{46}{9}, -\frac{10}{9} \right) \]